Problem: Find the value of $c$ so that $(x+1)$ is a factor of the polynomial $p(x)$. $p(x) = 5x^4+7x^3 -2x^2 -3x +c$ $c=$
Explanation: The following statements are equivalent: $(x+1)$ is a factor of $p(x)$ $p(x)$ is divisible by $(x+1)$ The remainder of $\dfrac{p(x)}{x+1}$ is $0$ We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $p(x)$ is divided by $(x+1)$, which can be rewritten as $(x-({-1}))$, is equal to $p({-1})$. We want this remainder to be equal to $0$. So let's set $p({-1})=0$ and solve this equation to find $c$. Let's plug ${x=-1}$ in $p( x) = 5 x^4+7 x^3 -2 x^2 -3 x +c$ and set that equal to $0$. $\begin{aligned} 5({-1})^4+7({-1})^3 -2({-1})^2 -3({-1})+c&=0 \\\\ 5-7-2+3+c&=0 \\\\ -1 +c&=0 \\\\ c&=1 \end{aligned}$ In conclusion, $c=1$.